15th Junior Turkish Mathematical Olympiad

Let $m^2=n^2+2025n+2010\cdot 15$ where $m$ is a positive integer. Then $(2n+2m+2025)(2n-2m+2025)=1995^2$. Hence $n=(A+B-4050)/4$ and $m=(A-B)/4$

where $A$ and $B$ are integers such that $AB=1995^2$, $A>B$ and $A+B>4050$. Since $1995^2\equiv 1(\mathrm{mod}4)$, these formulas always give integer values for $m$ and $n$. Let $f(x)=x+\frac{1995^2}{x}$. Then $f(x_2)-f(x_1)=(x_2-x_1)\left(1-\frac{1995^2}{x_1{x}_2}\right)>0$ for $x_2>x_1>1995$. The first divisor of $1995^2=3^2\cdot 5^2\cdot 7^2\cdot 19^2$ after $1995$ is $A=2025=3^2\cdot 5\cdot 7^2$ for which $f(A)>4050$ fails. The next one $A=2527=7\cdot 19^2$ and hence all the greater divisors satisfy $f(A)>4050$. The answer is $((2+1{)}^4-1)/2-1=39$.